Calculando a massa molar (molar peso) Para Calcular a massa molar de hum Composto químico, ponha SUA fórmula E clique em 'Calcular'. Na Fórmula química that rápido Você PODE USAR: <span>QUALQUÉR elemento químico Grupos Funcionais: D, Ph, Me, Et, Bu, AcAc, Para, Ts, Tos, Bz, TMS, tBu, Bzl, Bn, Dmg PARENTESIS () UO colchetes [] . Nomes Comuns de Compostos. </span><span>Os Exemplos de Cálculos de Massa molar: </span> NaCl <span>, </span> o Ca (OH) 2 <span>, </span> K4 [Fe (CN) 6] <span>, </span> CuSO4 * 5H2O <span>, </span> água <span>, </span> ácido nítrico <span>, </span> permanganato de potássio <span>, </span> etanol <span>, </span> frutose . Peso Computação molecular (massa molecular) para calcular o peso molecular de um composto químico entrar em sua fórmula, especifique seu número de massa isotópica depois de cada elemento entre colchetes. <span>Os Exemplos de Cálculos de peso molecular:</span> C [14] o [16] 2 <span>,</span> S [34] O [16] 2 <span>.</span> Definição de massa molecular, o peso molecular, uma massa molar e do peso molar <span><span><span>Massa molecular</span></span></span><span><span> ( </span></span><span><span><span>peso molecular</span></span></span><span><span> ) e uma massa de uma molécula de uma substancia e e Expressa nsa unificadas unidades de massa atómica (u). </span></span><span><span>(1 u E igual a 1/12 da massa de hum átomo de carbono-12) </span></span><span><span><span>Massa Molar</span></span></span><span><span> ( </span></span><span><span><span>molar peso</span></span></span><span><span> ) E uma massa de Uma toupeira de Uma substancia ê ê expresso em g / mol.</span></span><span> Pesos dos Átomos e isótopos São de </span><span>NIST Artigo </span><span>.</span> DEIXE Seu comentário <span>Sobre a SUA Experiência com uma calculadora de Peso Molecular </span><span>Pesos </span><span>moleculares de Aminoacidos: Relacionados</span>
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Answer:
The correct answer is control group.
Explanation:
A group used in a study or in an experiment, which does not get treatment by the scientists and is used as a foundation to determine the functions of the other tested subjects is known as the control group. The control group is only found in an experimental investigation.
The group in an experiment, which gets the variable being examined is known as an experimental group. The comparison of an experimental group is done with a control group in order to find the answers in an experiment.
Answer: b} The exact time when an individual atom will decay can be accurately predicted.
c} After each half-life, the amount of radioactive material is reduced by half.
Explanation:
All radioactive decay follows first order kinetics.
Rate law expression for first order kinetics is given by:
where,
k = rate constant
t = time taken for decay process
a = initial amount of the reactant
a - x = amount left after decay process
Expression for calculating half life, which is the time taken by the half of the reactants to decompose is:
Answer:
what are the answers?
Explanation:
Probably going to be c for this, until the answers you choose are shown. very very sorry :c
Answer:
Only changes in temperature will influence the equilibrium constant 
. The system will shift in response to certain external shocks. At the new equilibrium 
will still be equal to , but the final concentrations will be different.
The question is asking for sources of the shocks that will influence the value of . For most reversible reactions:
- External changes in the relative concentration of the products and reactants.
For some reversible reactions that involve gases:
- Changes in pressure due to volume changes.
Catalysts do not influence the value of . See explanation.
Explanation:
.
Similar to the rate constant, the equilibrium constant depends only on:
the standard Gibbs energy change of the reaction, and
the absolute temperature (in degrees Kelvins.)
The reversible reaction is in a dynamic equilibrium when the rate of the forward reaction is equal to the rate of the backward reaction. Reactants are constantly converted to products; products are constantly converted back to reactants. However, at equilibrium 
the two processes balance each other. The concentration of each species will stay the same.
Factors that alter the rate of one reaction more than the other will disrupt the equilibrium. These factors shall change the rate of successful collisions and hence the reaction rate.
- Changes in concentration influence the number of particles per unit space.
- Changes in temperature influence both the rate of collision and the percentage of particles with sufficient energy of reaction.
For reactions that involve gases,
- Changing the volume of the container will change the concentration of gases and change the reaction rate.
However, there are cases where the number of gases particles on the reactant side and the product side are equal. Rates of the forward and backward reaction will change by the same extent. In such cases, there will not be a change in the final concentrations. Similarly, catalysts change the two rates by the same extent and will not change the final concentrations. Adding noble gases will also change the pressure. However, concentrations stay the same and the equilibrium position will not change.
